Hi, dzatkovich-ga:
The chance of getting 6 or more pairs of Ace's in 150 random two-card
hands (from a standard card deck) is 1 minus the chance of getting 5
or fewer such hands.
The usual way to calculate this is then to use the binomial
distribution for independent repeated trials (Bernoulli trials). For
a single hand as you know:
Pr( hand is AA ) = (4/52)*(3/51) = (1/13)*(1/17) = 1/221
Pr( hand not AA ) = 1 - Pr( hand is AA ) = 220/221
Now compute the first six terms of the binomial expansion:
( Pr( hand is not AA ) + Pr( hand is AA ) )^150 = 1
and these terms correspond exactly to:
Pr( 0 of 150 hands are AA ) = (220/221)^150
= 0.50647856561905368169250739768815
Pr( 1 of 150 hands are AA ) = 150 * (220/221)^149 * (1/221)
= 0.34532629474026387388125504387829
Pr( 2 of 150 hands are AA ) = C(150,2) * (220/221)^148 * (1/221)^2
= 0.11694004071886208456433409440424
Pr( 3 of 150 hands are AA ) = C(150,3) * (220/221)^147 * (1/221)^3
= 0.026222918221805437144729463593678
Pr( 4 of 150 hands are AA ) = C(150,4) * (220/221)^146 * (1/221)^4
= 0.0043804192938697718866763990321257
Pr( 5 of 150 hands are AA ) = C(150,5) * (220/221)^145 * (1/221)^5
= 0.0005814011062772606322315947806276
Because these events are mutually exclusive, the probability of any
one occurring is the sum of all six probabilities:
Pr( 5 or less of 150 hands are AA ) = 0.99992963970013210980173399
The complementary probability is then:
Pr( 6 or more of 150 hands are AA ) = 0.000070360299867890198266
or roughly 7 times in 100,000.
regards, mathtalk-ga |